Primitive of Arctangent Function/Proof 2

Theorem

$\ds \int \arctan x \rd x = x \arctan x - \frac {\map \ln {x^2 + 1} } 2 + C$

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

$\ds u$

$=$

$\ds \arctan x$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac 1 {x^2 + 1}$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \arctan x \rd x$

$=$

$\ds x \arctan x - \int x \paren {\frac 1 {x^2 + 1} } \rd x + C$

$\ds $

$=$

$\ds x \arctan x - \int \frac {x \rd x} {x^2 + 1} + C$

rearranging

$\ds $

$=$

$\ds x \arctan x - \paren {\frac 1 2 \ln \paren {x^2 + 1} } + C$

$\ds $

$=$

$\ds x \arctan x - \frac {\ln \paren {x^2 + 1} } 2 + C$

simplifying

$\blacksquare$